Dominant root of a family of polynomials

Question

Let $f(x)=x^5-x^4-x^3-x^2-x-c$, where $c>2$ is a real number. It is easy to prove that there exists a positive real root $\alpha>2$ of $f(x)$ and all the other roots are non real.

Also, by using Kakeya-Enestrom theorem, it is possible to prove that if $\beta$ is a non real root of $f(x)$, then $|\beta|\leq \alpha$.

However, I would like to prove that this inequality is strict, but I am not being able to prove this. Someone has some suggestion?

The same problem happens by defining $f_c(x)=x^k-x^{k-1}-\cdots - x- c$, where $c>2$. So, $f_c(x)$ has only one positive root, say $\alpha$, which must be dominant, i.e., if $\beta$ is another root, then $|\beta|<\alpha$.

Thanks in advance!

Answer 1

Write the equation in the form $1=x^{-1}+x^{-2}+x^{-3}+x^{-4}+cx^{-5}:=f(x) $. If $|\beta|\geqslant \alpha$, the RHS has absolute value at most $f(\alpha) =1$ with equality if and only if $|\beta|=\alpha$ and all five summands $\beta^{-1} $ etc are positive reals. That is, $\beta=\alpha$.

Answer 2

$f(x)$ is the characteristic polynomial of its companion matrix $$ A = \pmatrix{0 & 0 & 0 & 0 & c\cr 1 & 0 & 0 & 0 & 1\cr 0 & 1 & 0 & 0 & 1\cr 0 & 0 & 1 & 0 & 1\cr 0 & 0 & 0 & 1 & 1\cr}$$ and $A^5$ has all entries $> 0$. Therefore by the Perron-Frobenius theorem there is a positive eigenvalue of multiplicity $1$ strictly greater in absolute value than all other eigenvalues.

This works for all $c > 0$.